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The Moon The Key, The Second A Natural Metric
By Ian Beardsley!
Copyright © 2023"
of 2 70
Contents
Abstract…………………………………………………………………..3
Foreward…………………………………………………………………4
1.0 Introduction…………………………………………………………5
2.0 The Mystery of the Moon and the Proton……………………..7
3.0 The Moon……………………………………………………………10
4.0 The Moon and the Livable Age of the Universe………………13
5.0 Sexagesimal…………………………………………………………21
6.0 Inertia As Proton-seconds……………………………………….24
7.0 The Proton Radius…………………………………………………28
8.0 The Proton Charge………………………………………………..39
9.0 Conclusion………………………………………………………….41
10.0 Conclusion 2………………………………………………………44
11.0 The Second Is The Metric………………………………………50
12.0 Kilogram-Seconds……………………………………………….55
13.0 Apophis…………………………………………………………….59
14.0 Solving The System………………………………………………62
15.0 Mars and the Moon………………………………………………66
Appendix 1………………………………………………………………69!
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Abstract
It turns out 1 second is not just Natural but the metric, it describes all the cycles of Universe in
its structural bases.!
We see the moon is not just connected to the protons that make elements but to the elements
gold and silver that are the metals for ceremonial jewelries and we see it is connected to when
the universe would die in the standard cosmological model, as well as it defines the unit of a
second as mysteriously natural. We develop the idea of inertia as proton-seconds which is
matter as a consequence of space and time and from it determine the radius of a proton. As
well the idea determines the charge of a proton. We see it arises from six-fold symmetries
which we suggest makes for a good basis of Nature.!
In this paper we ask why did we find the second to be Natural in terms of physics when it came
from the way the ancients made our calendar and divided it up. We find it is because their idea
was based on geometric principles that reconcile six-fold symmetry with eightfold symmetry,
and we include how this happens in the periodic table of the elements.!
It turns out 1 second is not just Natural but the metric, it describes all the cycles of Universe in
its structural bases. We see the life span of the universe is measured nicely by the second, as
well as the period of of one rotation of the the Earth, the earth day, and that the moon defines
the second as well as the radius of a proton and its mass."
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Foreward
One of things I wanted to go into in this paper but didn’t is that you can speak of the structure of
the solar system even though it changes with time.
The whole object of developing a theory for the way planetary systems form is that they meet the
following criterion: They predict the Titius-Bode rule for the distribution of the planets; the
distribution gives the planetary orbital periods from Newton’s Universal Law of Gravitation. The
distribution of the planets is chiefly predicted by three factors: The inward forces of gravity from
the parent star, the outward pressure gradient from the stellar production of radiation, and the
outward inertial forces as a cloud collapses into a flat disc around the central star. These forces
separate the flat disc into rings, agglomerations of material, each ring from which a different
planet forms at its central distance from the star (it has a thickness). In a theory of planetary
formation from a primordial disc, it should predict the Titius-Bode rule for the distribution of
planets. When I speak of the state of the solar system I speak of this point toward which the
solar system has formed and not the small changes that happen over millions of years due to
mutual interference between the bodies. In fact, mutual interference has torn apart possible
forming planets resulting in the current distribution we have today, because the current
distribution is more or less stable. The asteroid belt is a good example of this — it is a location
where a planet cannot form due to harmonic (repetitive) action on the orbital period at its
distance by orbital periods of planets beyond it. In short we take the state of the solar system as
an inflection point between what it became, and what it might minutely be going away from in
billions of years after it dies and can no longer support life.
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1.0 Introduction !
Our Natural satellite, the Moon, is surrounded by mystery, and after discovering something I
find intriguing, not that there is not already a lot that is intriguing, it has compelled me to outline
not just what I found, but what is already known, in hopes of answering some of the big
questions surrounding the mystery of life on Earth. I suspect the Moon may hold the key to
answering questions of why we are here, mainly because as seen from here on Earth, the Moon
perfectly eclipses the Sun, meaning that as seen from here it is about the same size as the
Sun, and since her path in the sky allows her to cover the Sun every so often that when we
witness a solar eclipse, we may be witnessing an intended display; but, from who or what we
obviously don’t know, but, the spectacle would have been witnessed as much by our early
ancestors sitting around a fire chipping stones into spearpoints 100,000 years ago as much as
by us today walking down the street in a town or city, or when hiking in the desert.!
The ecliptic is is the plane traced out by the Earth’s orbit to the Sun, and many of the planets in
our solar system are not too far out of this plane in their orbits, it is the line in the sky that the
Sun traces out over the course of a year, one orbit of the Earth around the Sun. The Earth
rotates on its axis at an angle of 23.44 degrees to the ecliptic. The equator of the Earth is
therefore tilted by this angle to its orbit, and the projection of the Earth’s equator onto the sky is
the celestial equator, parallel to which we see the stars rise and set because their apparent
motions are really due to the rotation of the Earth.!
The Moon is inclined to the ecliptic 5.15 degrees meaning since its ascending node to the
ecliptic rotates, its inclination to the Earth equator goes from !
18.3 deg = 23.44-5.15 to 28.6 deg= 23.44+5.15!
Where the ascending node is where the moon moves into the northern hemisphere of the
ecliptic and the descending node is where it moves into the southern hemisphere of the
ecliptic.!
The Moon orbits the earth with respect to the stars in 27.3 days, but with respect to the Sun in
29.5 days, because as it orbits the Earth, the Earth orbits the Sun, so with respect to the Sun it
chases the Earth’s rotation thus the period from New Moon to New Moon (It’s period in the
Earth sky) is a little longer than the sidereal month.!
It is believed the Moon makes life on Earth possible because it stabilizes the Earth’s angle of
rotation to the Sun. Without the Moon stabilizing the Earth’s orbital axis astronomers believe
the Earth’s tilt could vary as much as 85 degrees. Like this the sun would pass through the sky
nearly North to South and and change back to nearly East to West every few million years. Just
with the change the Earth has today of little more than a degree over thousands of years
causes the periodic ice ages we have. Astronomers are now beginning to think most terrestrial
planets around other stars do not have a Moon like ours stabilizing its rotation and hence life
sustaining seasons.!
The Moon is unique in that of the terrestrial planets it is the only one so closely spherical.
Which is because its mass is enough to pull all of its matter towards its center equally.!
When the solar system was first forming from the protoplanetary disk it was a chaotic place
with frequent bombardments upon the forming protoplanets. It is believed in this chaos, the
primordial Earth, not yet hardened but a soft blob, was hit by a Mars sized object ejecting
some of the mass of the Earth into orbit around the Earth from which the Moon formed. It is
calculated it had to be a Mars sized body that hit the Earth because only 10% of the matter
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ejected from Earth would have gone into the formation of the Moon. It is believed the moon’s of
the other planets did not actually form this way but were rather captured by the planets they
orbit.
Now on to what I found regarding the Moon, which is actually part of another work that is a
theory of reality (See Beardsley, The Mystery Of The Moon And The Proton, 2023). But in order
to proceed we need to compute the Kinetic energies of the Moon and Earth in their orbits.!
!
!
Using the Moon’s orbital velocity at aphelion, and Earth’s orbital velocity at perihelion. We also
want to list the values of the constants we will be using:!
(Proton Mass)!
(Planck Constant)!
(Proton Radius)!
(Gravitational Constant)!
(light speed)!
(Fine Structure Constant)!
!
!
(VanDerWaals Radius of Hydrogen)!
KE
moon
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
KE
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E33J
m
P
: 1.67262 × 10
−27
kg
r
p
: 0.833 × 10
−15
m
G: 6.67408 × 10
−11
N
m
2
kg
2
c: 299,792, 459m /s
α : 1/137
q
p
= q
e
= 1.6022E − 19coulombs
k
e
= 8.988E 9
Nm
2
C
2
R
H
= 1.2E − 10m
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2.0 The Mystery of the Moon and the Proton
We have the following two equations, one for the macrocosmos (Moon, Earth, Sun) and the
other for the microcosmos (proton):!
Eq 2.1. !
Eq. 2.2 !
Because equation 2.1 equals 1 second in the SI system of units (which is the mystery we want
to explore} it is a great basis for working with Nature, because it is composed of the
fundamental Natural Constants. We want to understand this equation that predicts the radius
of a proton within experimental errors, by using my equation for 1 second:!
Eq 2.3 !
Where!
!
Where is the golden ratio, and is Euler’s number equals 2.718…And is the
radius of a hydrogen atom, we get for the radius of a proton!
Eq. 2.4. !
Where!
, , !
Where is a variable, the number of protons in multiplied by Avogadro’s number over
grams making these three equations, equations of state for the periodic table of the elements
in that for example if , the element, is set at the element carbon then!
, , !
Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a
molar mass of 12. We have 12-6=6 grams of protons in the 12 grams of protons and neutrons.!
1
6α
2
⋅
r
p
m
p
h4π
Gc
= 1secon d
KE
moon
KE
earth
(EarthDay) = 1secon d
R
H
πα
2
m
2
p
⋅
2
6
⋅
1
N
A
𝔼
⋅
h
G
= 1.067secon d s
2
6
≈
1
Φ(1 + Φ)
≈
e
Φ
Φ =
5 + 1
2
e
R
H
r
p
=
1
m
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
N
A
=
Z ⋅ 6E 23proton s
Z ⋅ gra m s
𝔼 =
Z ⋅ gra m s
Z ⋅ proton s
N
A
⋅ 𝔼 = 6E 23
N
A
𝔼
Z
𝔼
ℂ
ℂ =
6gra m s
6protons
N
A
=
6(6E 23proton s)
6gra m s
N
A
ℂ = 6E 23
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In this sense Avogadro’s number is seen to be a natural constant not just a convenient number
used to describe a large number of atoms or compounds. But we wish to show that the
second has coincidentally come out to be a Natural Unit given we got it from dividing the Earth
day into 24 units we called hours, that into 60 units we called minutes, and that into 60 units
we called seconds. It is Natural I am saying because we have equation 2.1 and 2.2:!
!
!
We want to factor 2.1 into a ratio between kinetic energies times a time, as we have in 2.2 for
the moon and earth, so we can unravel the mystery of the moon and the proton. We find the
second is Natural to the biological as well and that everything here involves a mystery of six-
fold symmetry. We have that!
Eq 2.5. !
Eq 2.6. !
From which instead of saying the left sides of these equations are seconds, we say they are
proton-seconds by not letting cancel with the bodies of these equations on the left, but
rather divide into them, which are in units of mass, giving us a number of protons. I say this is
the biological because as we shall see our equations are based on one second is 6 protons is
carbon, and 6 seconds is one proton is hydrogen, these making the hydrocarbons which are
the skeletons of biological life. We see this is a mystery of six-fold symmetry based around
biological life in the following computer program I wrote and its output:!
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by
the mass of a proton. But these masses can be considered to cancel and leave pure number.
We make a program that looks for close to whole number solutions so we can create a table of
values for problem solving.
1
6α
2
⋅
r
p
m
p
h4π
Gc
= 1secon d
KE
moon
KE
earth
(EarthDay) = 1secon d
1
α
2
m
p
h4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h4π r
2
p
Gc
= 1proton ⋅ 6secon d s = hydrogen(H )
m
p
1
α
2
m
p
h4π r
2
p
Gc
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By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest
of the elements heavier than carbon are formed by fractional seconds. These are the
hydrocarbons the backbones of biological chemistry. Here is the code for the program:
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}!
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3.0 The Moon
Essentially that the Moon perfectly eclipses the Sun as seen from the Earth means while it is
400 times smaller than the Sun it is 400 times further from the Sun than it is from the Earth.
This determines its orbital velocity and mass, as well as that of the Earth and the mass of the
Sun. The orbital velocities of the Moon and the Earth are given by:!
Equation 3.1. and !
Equation 3.2. !
!
The Moon perfectly eclipses the Sun because!
Equation 3.3. !
Where is the Earth orbital radius, is the lunar orbital radius, is the solar radius, and
is the lunar radius. This gives:!
Equation 3.4. !
We have more explicitly:!
Equation 3.5. !
!
The rotational period of the moon is and the orbital
period of the Earth is is 27.3 which gives!
Equation 3.6. !
v
e
=
GM
⊙
r
e
v
m
=
GM
e
r
m
v
e
v
m
=
M
⊙
M
e
⋅
r
m
r
e
M
⊙
M
e
⋅
r
m
r
e
=
1.989E30kg
5.972E 24
⋅
1.74E6m
6.957E8m
= 28.86
r
e
r
m
≈
R
⊙
R
m
r
e
r
m
R
⊙
R
m
v
e
v
m
=
M
⊙
M
e
⋅
R
m
R
⊙
= 28.6
v
e
v
m
=
R
m
R
⊙
M
⊙
M
e
⋅
r
e
r
m
R
m
R
⊙
M
⊙
M
e
⋅
r
e
r
m
=
1.74E6m
6.957E8
1.989E30kg
5.972E 24kg
⋅
1.496E11m
3.84748E8m
= 28.46
T
m
= 27.3days = 2,358, 720secon ds
Γ
e
= 24hours = 86,400secon d s
T
m
Γ
e
=
R
m
R
⊙
M
⊙
M
e
⋅
r
e
r
m
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Thus while second is given by:!
!
The Lunar month is given by!
Equation 3.7. !
We also found in terms of the proton, the second is given by!
!
Which gives the radius of the proton as!
!
Where is one second approximated by!
!
Which gives the radius of a proton is!
!
The lunar month with respect to the stars (sidereal month) is 27.3 days. And with respect to the
Sun (synodic month) is 29.53059 days. So we are in the right area. We have suggested the
second is a Natural unit, but we might say the month is as well. In a sense we already knew
this because we have always known the Moon perfectly eclipses the Sun, and here we have
shown that is determined by the mass of the Earth, the radius of the Sun, its mass, and these
determine the Moon’s orbital distance, size, and mass, as well the Earth’s mass. The Moon
also has a function; it allows for life on Earth because its orbit holds the Earth’s inclination to its
orbit which allows for the seasons. !
The Sun’s radius to the lunar orbital radius is the ratio of the molar mass of gold to that of silver
which is about 1.8, or nine-fifths. The Sun is gold in color and the Moon is silver in color. We
have:
KE
moon
KE
earth
(EarthDay) = 1.08secon d s
T
m
=
R
m
R
⊙
M
⊙
M
e
⋅
r
e
r
m
⋅ Γ
e
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon ds
r
p
=
18
3
α
2
m
p
Gc
4πh
t
1
= 8.288587 × 10
−16
= 0.829f m
t
1
R
H
16πα
2
m
2
p
⋅
3 2
N
A
𝔼
⋅
h
G
= 1.12secon ds
r
p
=
9
8
2 ⋅
1
m
p
hc
4π
3
G
⋅
R
H
N
A
𝔼
SolarRa dius
Lu n ar Orbit
=
6.957E8m
3.844E8m
=
Au
Ag
=
196.97
107.87
= 1.8
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Considering equation 3.7 and this fact, we have:
Putting the values into 3.7 we have:
Converting this to days: 2459099/60/60/24=28.46 days
Remember the sidereal month is 27.3 days, and the synodic month is 29.53059 days. We can
write 3.7 in terms of the mass of a gold atom to the mass of a silver atom:
Equation 3.8.
2435130.93/60/60/24=28.184s
This is interesting, because Au/Ag is not exactly 9/5 but is 1.82599 it adjusts the Lunar month in
such a way that it is nearly 28 days even which gives 12.95848 which would be an almost exactly
13 month year that works better than our 12 month year. January has 31 days, February 29,
March 31, April 30 May 31, June 30, July 31, August 31, September 30, October 31, November
30, December 31. The average of the 29, 30, 31 day month is
Which is probably why we have 29, 31, 30 day months even though the synodic month (period
from new moon to new moon) is 29.53049 days, because 30 is evenly divisible into the 360
degrees of a circle and into the 60 of the sexagesimal counting system of the Sumerians,
Babylonians, and Ancient Greeks from which our calendar emerged and from which we get 60
seconds in a minute, and 60 minutes in an hour.
T
m
=
R
m
R
⊙
M
⊙
M
e
⋅
r
e
r
m
⋅ Γ
e
R
⊙
r
m
=
6.957E8m
3.844E8m
=
Au
Ag
=
196.97
107.87
= 1.8
T
m
=
1.74E6m
6.957E8m
1.989E 30kg
5.972E 24kg
⋅
1.496E11m
3.84748E8m
86400s = 2459099s
T
m
=
R
m
r
m
Au
Ag
M
⊙
M
e
⋅
r
e
r
m
⋅ Γ
e
T
m
= R
m
Ag
Au
M
⊙
M
e
⋅
r
e
r
3
m
⋅ Γ
e
T
m
= 1.74E6m
107.87
196.97
1.989E 30kg
5.972E 24kg
⋅
1.496E11m
(3.84748E8m)
3
86400s = 2435130.93s
31 + 29 + 30
3
=
90
3
= 30
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4.0 The Moon and the Livable Age of the Universe
Warren Giordano wrote in his paper The Fine Structure Constant And The Gravitational
Constant: Keys To The Substance Of The Fabric Of Space, March 21, 2019:
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where
‘ ’ is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational
constant numerically, but neglecting any units.
I found I could eliminate the and at the same time get the six of the six-fold symmetry with
which I was working by considering Avogadro’s number .
I suggested there exists some k that serves as a constant that describes both the microcosmos
and macrocosmos from the proton, to the atoms, to planetary orbits. It is such that the square
root of it times the earth orbital velocity is 6, because we are guessing we are dealing with six-
fold symmetry as the basis of Nature. That is
Eq. 4.1
We have that k is
Eq. 4.2
This follows from what Warren Giordano noticed that
Eq. 4.3
Without the right units. I noticed since Avogadro’s number is that I
could introduce an equation of state for the periodic table of the elements:
Eq. 4.4
Eq 4.5
Let us say we were to consider Any Element say carbon . Then in general
We have
and
h
1 + α
α
10
23
10
23
6.02E 23atom s
k v
e
= 6
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
h(1 + α) ⋅ 10
23
= G
6.02 × 10
23
≈ 6 × 10
23
ℍ = 1
gr a m
atom
h
(1 + α)
G
N
A
H = 6.0003
kg
2
⋅ s
m
𝔼
ℂ
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
ℂ =
6gr a m s
6protons
N
A
=
6(6E 23protons)
6gr a m s
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Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a
molar mass of 12. We have
12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus
And it follows that
We see in general since the atomic number Z is the number of protons in an atom that in general
this holds for all elements because
And
Therefore we always have:
is a variable, the number of protons in multiplied by Avogadro’s number.
Put in the Earth mean orbital velocity which is 29.79km/s (Zombeck, Martin V. 1982). We get:
Equation. 4.6
While we have masses characteristic of the microcosmos like protons, and masses characteristic
of the macrocosmos, like the upper limit for a star to become a white dwarf after she novas (The
Chandrasekhar limit) which is 1.44 solar masses — More mass than that and she will collapse —
we do not have a characteristic mass of the intermediary world where we exist, a truck weighs
several tons and tennis ball maybe around a hundred grams. To find that mass let us take the
geometric mean between the mass of a proton and the mass of 1.44 solar masses. We could take
the average, or the harmonic mean, but the geometric mean is the squaring of the proportions, it
is the side of a square with the area equal to the area of the rectangle with these proportions as
its sides. We have:
Equation. 4.7
We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is .
We have the intermediary mass is:
Equation 4.8
N
A
ℂ = 6E 23
h
(1 + α)
G
⋅ N
A
ℂ = 6.0003kg
2
⋅
s
m
𝔼
N
A
=
Z ⋅ 6E 23pr oton s
Z ⋅ gr a m s
𝔼 =
Z ⋅ gr a m s
Z ⋅ proton s
N
A
⋅ 𝔼 = 6E 23
N
A
𝔼
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 422.787kg
M
⊙
= 1.98847E 30kg
m
p
= 1.67262E − 27kg
m
i
= (2.8634E 30)(1.67262E − 27) = 69.205kg
of 15 70
All we really need to do now is divide equation 4.6 by equation 4.8 and we get an even number
that is the six of our six-fold symmetry.
Equation. 4.9
The six of our six-fold symmetry.
We have something very interesting here. We have
This is:
Eq. 4.10
Where k is a constant, given
Eq. 4.11
It was the Indian physicist Chandrasekhar who found the limit in mass for which a white dwarf
will not have its gravity overcome the electron degeneracy pressure and collapse. The non-
relativistic equation is:
Let us approximate 0.77 with 3/4. Since we have our constant
Eq 4.12
Eq. 4.13
Then
Eq. 4.14
Since our constant k in terms the Chandrasekhar limit is
Eq. 4.15
1
m
i
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 6.1092 ≈ 6
1
69.205kg
6kg
2
s
m
v
e
= 6
k v
e
= 6
k =
1
773.5
s
m
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
k =
1
m
2
i
h
1 + α
G
⋅ N
A
𝔼
m
i
= Mm
p
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
ℏ = h /2π
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
of 16 70
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of
light squared, that is it represents the ground state. It is
Since
We are suggesting the earth orbit is the ground state for our planetary system. We suggest it
holds for any planetary system because k as we will see is a natural constant that solves many
physical problems on many levels, not just planetary systems but atomic systems and the
particles that make them up.
Let us now recall equations 2.5 and 2.6
While we have considered them to be proton-seconds because they are a mass divided by the
mass of a proton, we can consider these two masses to cancel and say they are equal to 1 second
and six seconds respectively. We have that carbon, which is to evaluate them at one second, is
the radius of a proton:
Eq. 4.16
This gives the radius of a proton is:
Eq 4.17.
Where . The experimental value of the proton radius is 0.833fm+/-0.014fm!
α
2
=
U
e
m
e
c
2
k v
e
= 6
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon ds = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6seconds = hydrogen(H )
1
α
2
m
p
h 4π r
2
p
Gc
= 6secon d s
r
p
=
18
3
α
2
m
p
Gc
4πh
t
1
= 8.288587 × 10
−16
= 0.829f m
t
1
= 1secon d
of 17 70
We want to hone our constant k best we can. It, and our intermediary mass, which we want to
hone as well, are derived from the Chandrasekhar limit, and, being based on Thermal and
Statistical Physics, uses a lot of approximations. It is actual value will probably be arrived at very
accurately with incoming data in astronomy for White Dwarf Stars. Values are often given from
1.39 solar masses to 1.44 solar masses mostly due to ranges given from varying stellar
compositions which can vary in metallicity and so forth from star to star. The non-relativistic
estimate from Chandrasekhar is
We made the estimate 0.77 approximately 3/4. But using the exact value above for our
intermediary mass we have
We had estimated and using
Thus precisely
We have
We have honing our
Thus since we said with our estimate
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.44 ⊙
m
i
= Mm
p
= 0.77
c
3
ℏ
3
G
3
N
m
2
p
1/2
ℏ =
h
2π
m
i
=
3
2
c
3
h
3
8π
3
G
3
m
2
p
1/2
= 67.9943kg
3
2
= 0.8660
0.77 = 0.8775
0.8775
0.8660
= 1.01328
m
i
m
i
= (67.9943)(1.01328) = 68.897kg ≈ 69kg ≈ 70kg
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼 =
1
773.5
⋅
s
m
k =
1
(68.897kg)
2
(6.62607E − 34) ⋅
(1.007299)
6.67408E − 11
⋅ 6.02E 23 = 0.001268291s /m
of 18 70
Considering equation 4.1 of section 4.0 page 16, we see our value is still close to integer 6:
And since we have since suggested we should rather be using not the average orbital velocity of
Earth but that at aphelion, we have
Even closer to integer six. Let us return to equation 2.1 and 2.2:
Eq 2.1.
Eq. 2.2
We want to factor 2.1 into a ratio between kinetic energies times a time as we have in 2.2. We
write 2.1 as
And noticing
Which gives
Is
And we know
1
k
= 788.4626
m
s
k v
e
=
29790m s /s
788.46m /s
= 6.145748
k v
e
=
29300m s /s
788.46m /s
= 6.095986
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
K E
moon
K E
earth
(Ear th Day) = 1secon d
6α
2
r
p
m
4
p
⋅ Gc
4πh
=
m
p
1secon d
m
3
p
Gc
h
= K E
36α
4
⋅
m
p
⋅ K E
4π
=
r
2
p
m
2
p
1secon d
K E =
4π
36α
4
⋅
r
2
p
m
p
t
2
1
t
1
=
1
6α
2
m
p
h 4π r
2
p
Gc
of 19 70
And have
We also notice that obviously by dimensional analysis
We have the kinetic energy and we need another kinetic energy to form the ratio:
We have the intermediary mass and the constant k inverted. We can write
This is
=3.03E56 seconds
But we also formulated
4π
36α
2
⋅
(0.833E − 15)
2
t
2
1
⋅ (1.67262E − 27) = 1.413E − 49J
m
3
p
Gc
h
=
4π
36α
4
⋅
r
2
p
m
p
t
2
1
h
G
⋅
c
3
m
p
= K E
h
G
⋅
c
3
m
p
(K E
moon
)
= 4.6654
K E = m
3
p
Gc
h
m
3
p
⋅
Gc
h
K E
(t im e) = aT im e
m
i
K E = m
i
(
1
k
)
2
K E = (68.897kg)(788.4626m /s)
2
= 42,831,358Joules
K E
1
K E
2
(t im e) = (aT im e)
m
i
(
1
k
)
2
m
3
p
Gc
h
(1secon d ) =
(42,831,358J )
(1.4130E − 49J )
(1secon d )
of 20 70
So we can write
1 Earth year = (365.25)(24)(60)(60)=31557600 seconds
The universe is theorized by standard models to die in 100 trillion years, which is when the last
stars born will die out. This is exactly 1E14 years
We see the moon is connected to how old the universe is theorized to become.
6.62607E − 34
6.67408E − 11
⋅
299792459
3
1.67262E − 27
= 1.599298E 29J
h
G
⋅
c
3
m
p
= 1.599298E 29J
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
(1secon d ) =
1.599298E 29J
42831358J
= 3.734E 21seconds
3.734E 21s
31557600s
= 1.1832332E14years ≈ 1E14years
of 21 70
5.0 Sexagesimal
It was Buckminster Fuller who said “Nature employs 60 degree coordination”. Sixty degrees
are the degrees of an equilateral triangle. The triangle is the structure that encloses an area
with the least amount of sides. Buckminster Fuller said, systems of triangles are the the only
inherently stable patterns. The regular hexagon, a six-sides polygon tessellates as equilateral,
60 degree, inherently stable equilateral triangles. It was the scientist Shubnikov who said
among the living organisms the pattern with which we most frequently meet is five-fold
symmetry. It is well known that the physical, like snowflakes, are six-fold symmetry. Thus we
should not be surprised that our Abstract Cosmology is founded on six-fold symmetry. Two
and three are the smallest prime numbers and their product is six. Two times six is twelve, the
number most evenly divisible by whole numbers for its size (it is a so-called abundant number:
divisible evenly by 1,2,3,4,6, their sum is 16 which is greater than twelve itself and, three times
six is eighteen, the cyclical Nature of the periodic table (18 groups). !
At the dawn of civilization, the Sumerians who settled down from wandering, gathering, and
hunting, to invent agriculture and and build ceramic homes, developed the first mathematics,
and it was this sexagesimal (base 60) that passed on to the Babylonians, then ended up with
the Ancient Greeks, who divided not just the hour into 60 minutes of time and the minute into
60 seconds of time, but who divided the sky into hours, minutes, and seconds of arc which
was measured in time counted by the rotation of the Earth, and its orbital period around the
Sun, and the orbital period of the Moon around the Earth.!
We are suggesting nature is founded on six-fold symmetry, six which is constructed by
multiplying together the first two prime numbers, 2 and 3. Two factorial is two, three factorial is
6. Two times six is twelve, the number of months in a year, approximated by our moon’s
approximately 12 orbits around the earth in the time it takes the earth to go around the Sun
once. There are four weeks in a lunar month, and!
!
Is the number of seconds in the 24 hour Earth day. The duration of one second comes from
dividing up time and the sky like this in base 60, that came from the Sumerians, Babylonians,
and Ancient Greeks and it is believed they did this because 60 is evenly divisible by!
1,2,3,4,5,6,…10,12,15,20,30,60,…!
The moon orbits the Earth in approximately 30 rotations of the Earth, hence the 30 day month.
January has 31 days, February 29, March 31, April 30 May 31, June 30, July 31, August 31,
September 30, October 31, November 30, December 31. This averages out to!
!
We divide a circle into 360 units called degrees, which is six squared times ten. As such the
equilateral triangle has each angle equal to 60 degrees. There are approximately 360 earth
rotations in the time it orbits the sun once, Hence in one day the Earth moves through
approximately one degree in one day in its 365 day journey around the Sun. The sidereal
month, the time it takes the moon to return to the same position against the background of the
stars is 27 days 7 hours 43 minutes. We see the power of sexagesimal (base 60) in computing
time with the month divided into days, hours, and seconds where there are 60 minutes in an
hour and 60 seconds in a minute because we can convert this easily into seconds with such a
system as here, the duration of this sidereal month in seconds:!
1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 60
2
= 86,400
31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31
12
=
367
12
= 30.58333
of 22 70
!
The length of the synodic month, the completion of the phases of the Moon, or the time it takes
the Moon to return to the same position with respect to the Sun, is 29 days 12 hours 44
minutes 3 seconds. It is a little longer than the sidereal month because the earth has moved
with respect to the sun by the time the moon has come back around to a full orbit. We have:!
!
We looked at the arithmetic mean of the days in the month over a year, Let us look at the
harmonic mean:!
=!
!
Gathering like terms in the denominator:!
=0.225806+0.034482759+0.133333=0.39362!
!
Which is the most frequent value for the days in a month. Now let’s look at the geometric mean
which tempers the various values with one another.!
=30.493!
But perhaps it is more telling to take the mean between the individual days of the month that
we use:!
!
!
0.032258+0.034482759+0.033333=0.10007!
!
!
We see that the arithmetic mean gives us exactly a 30 day month, hence we always speak
loosely saying a month is 30 days. !
(24 ⋅ 27 + 7) ⋅ 60
2
+ 43 ⋅ 60
1
+ 0 ⋅ 60
0
= 2,358,000 + 2,580 + 0 = 2360580sec /month
(24 ⋅ 29 + 12) ⋅ 60
2
+ 44 ⋅ 60
1
+ 3 ⋅ 60
0
= 2,548,800 + 2,640 + 3 = 2551443sec /month
12
1
31
+
1
29
+
1
31
+
1
30
+
1
31
+
1
30
+
1
31
+
1
31
+
1
30
+
1
31
+
1
30
+
1
31
12
0.032258 + 0.034482759 + 0.032258 + 0.033333 + 0.032258 + 0.033333 + 0.032258 + 0.032258 + 0.033333 + 0.032258 + 0.033333 + 0.032258
12
0.39362
= 30.486
12
31 ⋅ 29 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 30 ⋅ 31
31 + 29 + 30
3
=
90
3
= 30
3
1
31
+
1
29
+
1
30
3
0.10007
= 29.799 ≈ 30
3
31 ⋅ 29 ⋅ 30 = 29.98888 ≈ 30
of 23 70
Indeed when Buckminster Fuller speaks of Nature employing 60 degree coordination he is
speaking of the equilateral triangle as the fundamental unit of one for area. That is the six-sided
regular hexagon can be divided into six equilateral triangles, giving it an area of six. Indeed he
takes the equilateral triangle as one because not only does it enclose an area with the fewest
sides, it has the smallest area for its perimeter; a regular hexagon with a perimeter the same
size encloses more area. I would like to say the Ancient Sumerians, Babylonians, and Greeks
made this connection, because in their sexagesimal (base 60) system of notation, one is a
straight line with a triangle on top. See fig 1. Ancient Mesopotamian Cuneiform for writing. It is
not sexagesimal notation in the strict sense, as they have a unique symbol for a tens column,
but it was this sexagesimal structure that lead to the 60 minute hour and 60 second minute, as
well as the angular division projected onto the sky in 60 minutes of arc in a degree and 60
seconds of arc in a minute as a way of mathematically defining time dynamically in terms of the
motions of the Earth and moon.!
Fig. 1
of 24 70
6.0 Inertia As Proton-Seconds
We describe reality as being based in the six-fold. I really think Nature employs 6-fold
symmetry because I think it is the most dynamic being the product of the two smallest prime
numbers 2 and 3. Those are the smallest factors down to which anything can be reduced, so that
is why I think 6-fold is the basis of Nature, that the basis has to be reduced to the smallest
factors. As such we have 2x3=6, 3x3=9, 2x9=18 and 3x6=18. The periodic table of elements is
periodic over 18 groups, which means when you count to 18 the elements start over making
groups where they have similar properties falling into the same groups. These groups are
determined by electron configurations, for instance carbon is in group 14 which means since it
wants to have noble gas electron configuration, it gains 4 electrons to be the same as the noble
gas in group 18 of the same period. 18-14=4 meaning it ionizes as C4- so it has 4 electrons to
combine with 4 hydrogen atoms that are each H+ or so we can have two H atoms that can
combine with one oxygen atom which is O2- because oxygen is in group 16 meaning 18-16=2,
and so forth making life possible. We now need to show that the fundamental particles that
build reality are based on sixfold symmetry for their mass, size, and charge and we want to do it
in terms of gravity on the macroscale, thus it has to use G the universal constant of gravitation
and Planck’s constant h, that quantizes energy on the microscale. I find I can do this as such:
The natural constants measure the properties of space and time. We can list some of them here:
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
Gravity is a property of space measured by the universal constant of gravity, G:
Equation 6.1
Matter, or inertia, which measures matter’s ability to resist a force is for each particle (protons
and neutrons) we will suggest given by:
Equation 6.2
Which describes mass per meter over time, which is:
Equation 6.3
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G:6.67408 × 10
−11
N
m
2
kg
2
c : 299,792, 459m /s
α : 1/137
q
p
= q
e
= 1.6022E − 19coulom bs
k
e
= 8.988E 9
Nm
2
C
2
G = 6.674E − 11N
m
2
s
2
= 6.674E − 11
m
3
s
2
⋅ kg
h
Gc
=
kg ⋅ s
m
1.82E − 16
kg ⋅ s
m
of 25 70
It must be adjusted by the fine structure constant . It is my guess the factor should be which
is 18,769.:
Equation 6.4
Because the fine structure constant squared is the ratio of the potential energy of an electron in
the first circular orbit to the energy given by the mass of an electron in the Bohr model times the
speed of light squared, that is it represents the ground state. It is
Equation 6.5
Here we are suggesting that the proton and neutron are the 3-dimensional cross-sections of a
hypersphere. Thus we consider the surface area of a proton, :
Equation 6.6
We take the square root to get meters:
Equation 6.7
We multiply that with the value we have in equation 6.4:
Equation 6.8
We find that the mass of a proton realizes if we divide this by six seconds:
Equation 6.9
That is hydrogen. We see that the element carbon manifests if we divide instead by 1 second:
Equation 6.10
Carbon (C) is the core element of life chemistry and it combines with hydrogen (H) to make the
skeletons of organic matter, the so-called hydrocarbons:
Equation 6.11
Equation 6.12
α
1
α
2
(1.82E − 16kg ⋅ s /m)(18,769) = 3.416E − 12kg ⋅ s /m
α
2
=
U
e
m
e
c
2
S
p
S
p
= 4π r
2
p
= 8.72E − 30m
2
S
p
= 2.953E − 15m
(2.953E − 15m)(3.416E − 12
kg ⋅ s
m
) = 1.009E − 26kg ⋅ s
m
p
m
h
=
1E − 26
6secon d s
= 1.67E − 27kg
m
c
=
1E − 26kg ⋅ s
1secon d
= 1E − 26kg = 6pr oton s = 6m
p
m
p
=
1E − 26kg ⋅ s
6secon d s
m
c
=
1E − 26kg ⋅ s
1secon d
of 26 70
It seems the duration of a second is natural. If it is, since it was formed by a
calendar based on reconciling the periods of the moon and the sun in the earth
sky, it should be in the Earth-moon orbital mechanics. I find it is, that (See Appendix 1):
Equation 6.13
That is, the earth day (86,400 seconds) times the kinetic energy of the moon to the kinetic
energy of the earth is about 1 second (about 1.2 seconds). The earth day changes very little, by
very small amounts over millions of years. The solar system has evolved towards this since the
explosion of life called the Cambrian, and will slowly decay away from it. But we need to derive
the second in terms of something else. For now we have the mass of a proton as:
Equation 6.14
This way of looking at things is to say matter is that which has inertia. This means it resists
change in position with a force applied to it. The more of it, the more it resists a force. We
understand this from experience, but what is matter that it has inertia? In this analogy we are
suggesting a proton is a three dimensional bubble embedded in a two dimensional plane. As
such there has to be a normal vector holding the higher dimensional sphere in a lower
dimensional space. (See Fig. 1) Thus if we apply a force to to the cross-section of the sphere in
the plane there should be a force countering it proportional to the normal holding it in a lower
dimensional universe. It is actually a 4-dimensional hypersphere whose cross-section is a
sphere. This counter force would be experienced as inertia. We are defining inertia as proton-
seconds, the action of subatomic particles over time. Our reasoning above in one equation is:
K E
moon
K E
earth
(Ear th Day) ≈ 1secon d
m
p
=
3r
p
18α
2
4πh
Gc
Fig. 1
of 27 70
Equation 6.15
That is 1 second gives carbon. We find six seconds gives 1 proton is hydrogen:
Equation 6.16
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon ds = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6seconds = hydrogen(H )
of 28 70
7.0 The Proton Radius
Thus we have the radius of a proton is given by carbon by evaluating at one second:
But to get that we have to multiply by one second and we need one second in terms of the atom
for a theory of the proton. I find we can do that…
Substitute for to get
We have now introduced the radius of a hydrogen atom . It seems we have to
divide by two which I think is because we are looking at packing of atoms. This radius of the
hydrogen atom is the Van Der Waals radius, which is the closest distance between two hydrogen
atoms noncovalently bound. It is 120 pm. Divide that by ck where 1/k is our constant
And we find
We have our equation for the radius of a proton
We only need to multiply it by to have the right units, and we get!
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
t
6
=
1
α
2
m
p
h 4π r
2
p
Gc
t
6
=
r
p
α
2
m
p
h 4π
Gc
R
H
/2
r
p
t =
R
H
2α
2
m
p
h 4π
Gc
R
H
= 1.2E − 10m
R
H
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
1 + α
⋅
1
N
A
𝔼
≈
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
N
A
𝔼
t
ck
=
3 2
16
⋅
6.626E − 34
6.674E − 11
⋅
18769
(1.6726E − 27)
2
⋅
1.2E − 10
6.02E 23
⋅
1
π
= 1.12secon d s
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
2
p
⋅
R
H
N
A
𝔼
= 1.12secon d s
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
t /ck = 1secon d
of 29 70
Equation 7.1.
Then suggest we picked up 9/8 in approximations which is close to one anyway so we write
Equation 7.2.
We form constants:
And we have the Equation:
Equation 7.3
We can say that Avogadro’s number is not an arbitrary number because it is such that there are
twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12. We now want to add
to this section the equation of the mass of a proton.
If our equation is right and we put it into natural units then the product should be close to
one:
Let us start with the units with which we are working:
r
p
=
9
8
2 ⋅
1
m
p
hc
4π
3
G
⋅
R
H
N
A
𝔼
r
p
=
1
m
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
r
p
= 8.26935E − 16m ≈ 0.827f m
r
p
=
1
kg
(
kg
m
2
s
)
(
m
s
)
(
kg
s
2
m
3
)
R
H
N
A
ℍ
=
1
kg
(
kg
)
m
k =
hc
2π
3
G
= 6.93E − 9kg
R
H
N
A
𝔼
= 1.99E − 34m
r
p
m
p
= k
R
H
N
A
𝔼
m
p
=
1
r
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
r
p
m
p
G =
m
3
kg ⋅ s
2
h = kg ⋅
m
2
s
c = m /s
of 30 70
And convert these to proton-masses and proton-radii:
Now we find k in these units:
Thus we have:
!
We will say our theory is holistic and that we are describing the proton radius in terms of the
whole of which it is a part, namely, the radius of a hydrogen atom, more specifically the Van der
Waals radius, which is determined by hydrogen gas, or . It is done like this:
Johannes Diderik van der Waals (1873) described more than just Ideal gases, which are gases
that behave according to kinetic-molecular theory, he described real gases which don’t. His
equation then, The Van der Waals equation, is a modification of the Ideal Gas Law which is:
Which is quite obvious. If you increase the temperature T, then the volume of the gas is going to
increase, and if it doesn’t then the pressure will, which is inversely proportional to volume.
However for a Real Gas, he assumed the particles are hard spheres, cannot be compressed
beyond a limit, and at close proximity to one another they interact and have a volume around
them that excludes one another, that is they have walls. He said
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅ 1.67262E − 27kg(0.833E − 15m)
3
⋅ s = 193,131, 756
h = 6.62607E − 34kg ⋅
m
2
s
⋅
s
(0.833E − 15)
2
(1.67262E − 27kg)
= 5.71E 23
c =
(299,792, 459m /s)(1sec)
(0.833E − 15m)
= 3.6E 23
R
H
=
1.2E − 10m
0.833E − 15m
= 144,058
k =
hc
2π
3
G
= 6.93E − 9kg
k =
(5.71E 23)(3.6E 23)
2π
3
(193131756)
= 4E18pr oton − m a sses
r
p
m
p
= k
R
H
N
A
𝔼
r
p
m
p
=
(4E18)(144058)
(6E 23)
=
5.76E 23
(6E 23)
= 0.96 ≈ 1
H
2
PV = n RT
V
R
= V
I
− b
of 31 70
That is the volume of the real gas ( ) is equal to the volume of the ideal gas ( ) minus a
correction factor b. The volume of the particles is the number of particles ( ) times the volume
of one particle:
Thus there exists a sphere of radius 2r formed by two particles in contact where no other
particles can enter. It gives the correction factor
And the volume correction for n particles is
V
R
V
I
n
n
4
3
π r
3
b = (4)
4
3
π r
3
nb = 4n ×
4
3
π r
3
of 32 70
This is the volume correction to the Ideal Gas Law. The pressure correction says real gases
exhibit less pressure because their particles interact which is a net pulling by the bulk of
particles away from the container walls.
The reduction in pressure is proportional to by a factor a. We have for reduction of pressure
that
We substitute this into the Ideal Gas Law:
This can be written as a cubic
Which allows one to compute the critical conditions of liquefaction and to derive an expression
of the principle corresponding states. In the cubic form we have as the solution three volumes
which can be used for computing the volume at and below critical temperatures.
Thus the Van Der Waals radius is estimated
For hydrogen experimentally. Therefore with
We have described the derivation of radius of a hydrogen atom from the Van Der Waals
equations that we use to get
Where the Van Der Waals equations are
n
2
v
2
P
I
= P
R
+ a
n
2
V
2
(
P + a
n
2
V
2
)
(
V − n b
)
= n RT
V
3
−
(
b +
RT
P
)
V
2
+
a
p
V −
a b
P
= 0
(
π +
3
φ
2
)
(3φ − 1) = 8τ : π =
P
P
c
, φ =
V
V
c
, τ =
T
T
c
4
3
π r
3
w
=
b
N
A
r
3
w
=
3
4π
⋅
b
N
A
b = 26.61
cm
3
m ol
N
A
= 6.02E 23
r
w
= 1.0967E − 8cm = 1.0967E − 10 m
2
6
R
H
h
π α
2
m
2
p
GN
A
= 1secon d
of 33 70
So we can compute the radius of a proton from
Which needs to be multiplied through by one second to get
But in determining the radius of a proton from
We need to determine k in terms of the Chandrasekhar limit where k is
Where
(
P + a
n
2
V
2
)
(
V − n b
)
= n RT
V
3
−
(
b +
RT
P
)
V
2
+
a
p
V −
a b
P
= 0
4
3
π r
3
w
=
b
N
A
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
r
p
=
1
m
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
r
p
=
18
3
α
2
m
p
Gc
4πh
t
1
= 8.288587 × 10
−16
= 0.829f m
t =
R
H
2α
2
m
p
h 4π
Gc
t
ck
=
3 2
16
⋅
6.626E − 34
6.674E − 11
⋅
18769
(1.6726E − 27)
2
⋅
1.2E − 10
6.02E 23
⋅
1
π
= 1.12secon d s
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
1 + α
⋅
1
N
A
𝔼
≈
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
N
A
𝔼
k =
1
m
2
i
h
1 + α
G
⋅ N
A
𝔼
m
i
= (2.8634E 30)(1.67262E − 27) = 69.205kg
M
⊙
= 1.98847E 30kg
of 34 70
Where we multiply by 1.44, the Chandrasekhar limit to get 2.8634E30kg giving us our
intermediate mass using the Chandrasekhar limit as the upper limit for a mass, and
The mass of a proton for the lower limit. To have our equation k
We needed to use the Chandrasekhar limit
So we address now how that limit is derived…
The pressure of the outer shell of star balances with the outward pressure in the core of
the star (thermal pressure). Pressure is force per unit surface area thus…
is the mass of the core pulling in the mass of the shell and is the radius of the
core. The surface area of the star is that of a sphere, . We have
The thermal pressure countering the gravity is given by the ideal gas law PV=nRT (pressure
times volume of a gas such as hydrogen , which is all protons , is proportional to temperature.
The number of protons in the core is . We have
Where is the Boltzmann constant ( ). Since we must have
if the star is not to implode or explode
M
⊙
m
i
m
p
= 1.67262E − 27kg
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
P
gravit y
P =
F
A
F = m a = PA
P =
m a
A
m a = G
M
shell
M
core
r
2
core
M
core
M
shell
r
core
A = 4π r
2
core
P
gravit y
= G
M
shell
M
core
4π r
4
core
m
p
N
p
M
core
m
p
P
thermal
=
M
core
m
p
⋅
1
4
3
π r
3
core
k
B
T
core
k
B
1.380649E − 23J ⋅ K
−1
P
gravit y
= P
thermal
of 35 70
And we have the estimate for the temperature of the core of a star.
Fusion would not occur at the low temperature of a star like the Sun in that there would not be
enough energy for collisions, unless the potential Coulomb barrier can be overcome by quantum
mechanical tunneling. The collisions are given by the kinetic energy of the particles
. We have
The velocity v yields the minimum distance between protons as the De Broglie wavelength
Since the velocity is the root mean square velocity of the protons…
We have the temperature of the star is
This is another estimate. Since the mass of a star is its volume times its density
But for a star density varies with radius
If we take the derivative of both sides of the equation we have one of the equations of stellar
structure:
1.
The so-called conservation of mass equation. The force on the shell of the star is given by the
mass of the shell
k
B
T
core
=
1
3
GM
shell
m
p
r
core
1
2
m
2
p
v
e
2
4πϵ
0
r
min
=
1
2
m
p
v
2
λ =
h
m
2
p
v
v
rms
=
3k
B
T
mp
T
min
=
(
e
2
4πϵ
0
)
2
m
p
3π
2
h
2
k
B
m =
4
3
π r
2
ρ
4π
∫
r
0
r
2
ρ(r)dr
dm(r)
dr
= 4π r
2
ρ(r)
of 36 70
Again for there to be balance gravitation pressure equals thermal pressure:
2.
Another equation of the equations of stellar structure. The so-called equation of hydrostatic
equilibrium. This can be written
If the star is an ideal gas the density of the star varies as where for a
monatomic gas and then
In stellar dynamics we write
So that
The abundance of hydrogen and helium in the universe are approximately 75% and 24%,
respectively. Thus for every 4He2+ there are 12H+ and 2+12 free electrons. We have
Ionized hydrogen and helium have and for the Sun because of high metal
content. Finally stars can be approximated as blackbody radiators (purely radiate) and as such
pressure is given in terms of temperature (Temperature is proportional to radiation energy):
There are three kinds of pressures that can be generated by a star: gas pressure, radiation
pressure, or degeneracy pressure.
A type of star that is stable, that is prevented from collapse by degeneracy pressure, is a so-called
white dwarf star. They are the remnant of giant stars that have depleted the their fusion fuel and
thereby collapsed under gravity but are kept from collapsing into black holes by thermal
pressure due to motion of the particles alone. Interestingly, they still shine almost as bright as a
F
g
= − G
m(r)4π r
2
ρ(r)
r
2
dr
dρ(r)
dr
= − G
m(r)ρ(r)
r
2
d
dr
(
r
2
ρ(r)
dP(r)
dr
)
= − 4π Gr
2
ρr
PV
γ
= con sta nt
γ = 5/3
P ∝
1
V
γ
∝ ρ
5/3
N
V
=
ρ
μm
p
P
gas
=
ρ
μm
p
k
B
T
4 + 12
1 + 12 + 14
= 0.59
μ = 0.59
μ = 0.62
P
rad
=
4
3
⋅
σ
c
T
4
of 37 70
star on the main sequence even though they are not doing fusion. It was the Indian physicist
Chandrasekhar who found the limit in mass for which a white dwarf will not have its gravity
overcome the degeneracy pressure and collapse. The non-relativistic equation is:
There are many resources available that derive this and you can find it in any textbook on
astrophysics in the chapters dealing with stellar physics, and I will leave the treatment of the
derivation to those works.
Returning to our equation for one second in terms of the Earth and Moon orbital kinetic
energies:
We notice it would be exactly one second if the Earth day were shorter. And, indeed it was a long
time ago.
For our equation Earth day needs to be shorter. A long time ago it was; the Earth loses energy to
the moon. The days become longer by 0.0067 hours per million years. Our Equation
Is actually 1.2 seconds for KE of moon and KE or earth calculated with their average orbital
velocities. We will compute how long ago the Earth day was what it was needed to turn that 1.2
seconds into one second, and then how long ago the Earth day was what it was needed to turn
the 1.08 seconds into one second.
We have
24-20=0.0067t
t=597 million years
This was when the earth went through a dramatic change and there was a big explosion of life
(The Cambrian). This is when the Moon predicted the second as exactly 1. The dinosaurs went
extinct 65 million years ago giving small mammals a chance to evolve paving the way for
humans.
24-x=0.0067t
x=23.5645 hours
We say 20 hours + 3 hours is 0 hours + 3 hours since 20 hours is the zero of our cosmic
calendar:
M ≤ 0.77
c
3
h
3
G
3
N
m
4
p
= 1.41 ⊙
K E
moon
K E
earth
(Ear th Day) = 1.08secon ds
K E
moon
K E
earth
(Ear th Day) ≈ 1secon d
24h ours
1.2
= 20h ours
of 38 70
What is the next term?
20+3+0.57735+0.4714=24hours
Which bring us to today. Now the calculation for the 1.08 seconds that works with KE of moon
and KE earth calculated using aphelions and perihelions.
24-22.222=0.0067t
t=265.373 million years
This is the Permian period which was from 299 to 251 million years ago, it was at the end of the
Paleozoic Era which was followed by the Mesozoic Era. The distinction between the Paleozoic
and the Mesozoic is made at the end of the Permian to mark the largest mass extinction
recorded in Earth’s history.
Thus, to conclude, we write
And focus for a bit on . It takes the square which has diagonal to side is and divides that
up with regular hexagon which is six radii and six sides equal one another reconciling Euler’s
number e with golden ratio characteristic of five-fold symmetry, the regular pentagon. That is
Equation 8.6.
Now we show what this is geometrically and explore how we arrived at it in the following
pages…
3cos(0
∘
) +
2
3
cos(30
∘
) = di nosaur − ext i nct ion =
20hrs + 3hrs +
3
3
hrs +
2
3
=
24h ours
1.08
= 22.222h ours
2
6
R
H
h
π α
2
m
2
p
GN
A
= 1secon d
2
6
2
Φ
2
6
=
1
Φ(1 + Φ)
=
e
Φ
of 39 70
8.0 The Proton Charge
I construct the electric field as such (Fig. 2): is one component of . Events in
are through time with components and where c is the speed of light. The
electrons and the protons and in the presence of one another cross into giving
themselves acceleration in the and directions.
We have
We suggest
Equation 8.1
The position of changes in the space of and has travelled to a place in
time, as well. Since ct=meters there is a g such that . Thus since
x
ℝ
3
(x, y, z)
ℝ
3
(x, y, z)
ct
1
ct
2
q
e
q
p
ct
1
ct
2
x
−x
i
j
k
0 ct
1
0
0 0 ct
2
= (ct
1
− ct
2
)
i
··
x ∝ (ct
1
− ct
2
)
i
x
ℝ
3
(x, y, z)
t (0,ct
1
, ct
2
)
··
x = g(ct
1
− ct
2
)
i
g ⋅ m =
m
s
2
Fig. 2
of 40 70
and that is . Thus g is frequency squared ( ) and we suggest it is
derived from the separation between the charges and . That there is some velocity v such
that . We call upon our equation for 1 second:
Letting v= from our equation for k
And we have
Equation 8.2
We get
Equation 8.3 !
g =
1
s
2
=
1
t
2
1
s
2
⋅
m
s
⋅ s =
m
s
2
f
2
q
e
q
p
f =
(
v
x
)
r
p
m
p
h 4π
Gc
= 1secon d
(
v
x
)
2
(c)
r
p
m
p
h 4π
Gc
=
k
e
m
⋅
q
2
x
2
v
2
(c)
r
p
m
p
h 4π
Gc
= k
e
q
2
m
α
2
6
⋅
1
k
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼 =
1
773.5
s
m
1
k
= 773.5
m
s
m = m
p
c
k
e
(
α
2
6
⋅
1
k
)
2
h 4π r
2
p
Gc
= q
2
h 4π r
2
p
Gc
=
(6.626E − 34)(4π)(0.833E − 15)
2
(6.674E − 11)(299,792, 459)
= 5.37E − 31
q = (0.033)
1
36(18769
2
)
773.5
2
(5.37E − 31) = 9.1435E − 19C
q
q
p
= ⋅
9.1435E − 19C
1.602E − 19C
= 5.71protons ≈ 6protons
of 41 70
9.0 Conclusion
We would like to present some things clearly:
Thus
If is 1.00 seconds even. In terms of our theory
Or as another estimate:
Because
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
t =
R
H
2α
2
m
p
h 4π
Gc
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
1 + α
⋅
1
N
A
𝔼
≈
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
N
A
𝔼
t
ck
=
3 2
16
⋅
6.626E − 34
6.674E − 11
⋅
18769
(1.6726E − 27)
2
⋅
1.2E − 10
6.02E 23
⋅
1
π
= 1.12secon d s
r
p
=
18
3
α
2
m
p
Gc
4πh
t
1
= 8.288587 × 10
−16
= 0.829f m
t
1
t
1
=
R
H
π16α
2
m
2
p
⋅
3 2
N
A
𝔼
⋅
h
G
= 1.12secon d s
t
1
=
R
H
πα
2
m
2
p
⋅
2
6
⋅
1
N
A
𝔼
⋅
h
G
= 1.067secon d s
3 2
16
= 0.265165
2
6
= 0.23570
0.23570
0.265165
= 0.888888
of 42 70
Or, in the sense that
And the life span of the Universe is
Is 1E14 years or 100 trillion years. Given we have
Eq 2.1.
Eq. 2.2
Then
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
t
1
=
1
α
2
6m
p
h 4π r
2
p
Gc
t
6
=
1
α
2
m
p
h 4π r
2
p
Gc
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon ds = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6second s = hydrogen(H )
m
i
=
3
2
m
p
c
3
h
3
8π
3
G
3
m
2
p
k =
4
3
m
p
8π
3
G
c
3
h
⋅ N
A
𝔼
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
(1secon d ) =
1.599298E 29J
42831358J
= 3.734E 21secon ds
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
K E
moon
K E
earth
(Ear th Day) = 1secon d
of 43 70
Where
And
Where
The first k above can also be written
Our value for k can vary because the Chandrasekhar limit can range from 1.39 to 1.44
depending on the exact composition of a white dwarf star.
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
⋅
K E
moon
K E
earth
(Ear th Day) = Li feSpa nUniverse
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
⋅
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= Li feSpanUniverse
k v
e
= 6
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
1
k
= 788.4626
m
s
r
p
m
p
= k
R
H
N
A
𝔼
k =
hc
2π
3
G
= 6.93E − 9kg
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼 =
1
773.5
⋅
s
m
of 44 70
10.0 Conclusion 2
If the universe was born from a big explosion called “the big bang” then it is expanding and the
distances between galaxies is growing. This does not mean there is an edge to the universe —
space could continue on infinitely in all directions but there could be a horizon beyond which
there is no content.
There can be an observable edge to the universe, but it is more of a horizon, just as there is a
horizon to the ocean, we know beyond the horizon there is more water. Since the universe was
born 13.82 billion years ago in the Big Bang we can only see those galaxies whose light has taken
less than 13.82 billion years to reach us. These galaxies form a sphere around the Earth called
“the observable universe”. It is 92 billion light years in diameter because in its beginning the
universe inflated much faster than the speed of light.
The Schwartzchild radius:
r𝚜=2GM/c²
Gives the event horizon of a black hole. I have heard it said if the mass, M, is the mass of the
universe, that radius is the edge of the observable universe that, the observable edge of the
universe is actually the event horizon of a black hole and that then the Universe is a giant black
hole and that we are inside of it. I would like to call this to question, but first let’s discuss the
types of possible universes.
The pivotal parameter describing the universe is 𝛀 (omega) which is the average matter density
of the universe divided by a critical value of that density. Whether omega is less than1, 1, or
greater than 1 determines whether the Universe is open, flat, or closed. If matter is mostly inert
as in the dust models, there is a particular fate for each omega. Since 1998 the observations in
supernovas indicate the universe is accelerating in its expansion. To explain this cosmologists
hypothesize the existence of dark energy which can be any field with negative entropy causing
negative pressure. If 𝛀<1 then the Universe is open and the fate of the universe is in its heat
death, the burning out of the last stars born. This is also true of a flat universe if it has the
hypothesized dark energy, because then it accelerates to escape collapse. The fate of the universe
depends on its density and today most evidence points to it will expand indefinitely. In this
scenario there is enough a supply of gas for stars to form for 10¹² to 10¹⁴ years. 1 to 100 trillion
years. That is the last stars born would die in 10¹⁴ years. There are models that suggest the
universe is eternal where as stars die more matter comes into existence from which more stars
are continuously forming.
Now let’s call into question that the edge of the observable universe is the Schwartzchild radius,
the event horizon of a blackhole.
I had seen a documentary where they said there was a cosmologist working on a theory that the
edge of the observable universe is the event horizon of a blackhole. I thought okay as weird as
that seems, and I say weird because that means either the Earth is at the center of this blackhole,
or if not, mysteriously wherever you are in the universe is the center. I accepted this as possible
because the more we learn the stranger the universe gets. Indeed a long time ago Niels Bohr said
that if you haven’t realized that quantum mechanics says everything that makes up the universe
can’t be considered real, you have missed its point. And indeed if you think about quantum
mechanics, this rings true, so yes I was able to accept this about the edge of the universe, as
possible. However, taking a second look we see what is going on here may be circular
reasoning…
of 45 70
Einstein’s equations give the relationship between spacetime and matter content of the
Universe. From this Friedmann gave us his equations that are at the heart of cosmology, which
are functions of time, the scale-factor R, the matter density of the universe , and the pressure p
due to radiation produced by the stars, or by the galaxies because they contain the stars:
Where
for a 3-sphere, for 3-space, and for a 3-pseudosphere. These Friedmann
equations give the critical density of the universe as
Where is the Hubble constant which is the expansion rate of the universe. Current estimates
are that it is . A Hubble sphere is given, which is given by
,
Since the density of the universe is very close to the critical density which is
, then since
So this is close to the observable edge of the Universe. But the radius of the Universe is much
larger than 13.8 billion light years, because it is expanding we say it is 46.508 billion light years.
However we should look at where we got this mass we used, it came from the Hubble constant
and we used . What we did is:
ρ
2
R′ ′
R
+
(
R′
R
)
2
+
κ
R
2
+ 8π p = 0
(
R′
R
)
2
+
κ
R
2
=
8πρ
3
R′ ≡
dR
dt
κ = + 1
κ = 0
κ = − 1
ρ
c
=
3H
2
0
8π G
H
0
71k m /s /Mpc
R
HS
=
c
H
0
=
299792549m /s
71,000m /s /Mpc
= 4222.429Mpc
pc = 3.26ly
Mpc = 3.26E6l y
R
HS
= 13.8E 9ly
ρ
c
≈ 8.6E − 30g/cm
3
= 8.6E − 27kg/m
3
R
HS
= 1.38E10ly = 1.3E 26m
M = ρ
c
4
3
π R
3
= 8E52kg
R
s
=
2GM
c
2
= 2
(6.67E − 11)(8E52kg)
299792459
2
= 1.18742E 26m = 12.55E 9ly
ρ
c
of 46 70
Which is
Which is the critical mass given by the Friedmann equations. So, we have gone in a circle, I
think. Let us use our value of 46.508E9 ly = 4.4E26m instead of 1.3E36m:
That is larger by a factor of 36.255. Not to say we aren’t in a blackhole, it just can’t be at the edge
of the observable universe, the radius of the Hubble sphere. Now with this brief sketch of
cosmology that we have done, let us talk again about some of the conclusion in this paper. We
have the second may be a natural duration. We suggest this because we found!
!
Or…!
!
R
s
=
2GM
c
2
= 2
G
c
2
ρ
4
3
π
(
c
H
0
)
3
=
8π G
3H
2
0
ρ
c
R
ρ
c
=
3H
2
0
8π G
M = ρ
c
4
3
π R
3
= 3.0686E54kg
R
s
=
2GM
c
2
= 2
(6.67E − 11)(3.0686E54kg)
299792459
2
= 4.55E 27m = 4.81E11ly
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
K E
moon
K E
earth
(Ear th Day) = 1secon d
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
⋅
K E
moon
K E
earth
(Ear th Day) = Li feSpa nUniverse
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
⋅
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= Li feSpanUniverse
t
1
=
R
H
π16α
2
m
2
p
⋅
3 2
N
A
𝔼
⋅
h
G
= 1.12secon ds
t
1
=
R
H
πα
2
m
2
p
⋅
2
6
⋅
1
N
A
𝔼
⋅
h
G
= 1.067secon d s
of 47 70
But let us ask why did the unit of a second come out as natural in the structure of Nature and
the Universe. We got it from ancient times by dividing the Earth day (1 rotation of the Earth) into
24 hours and that into 60 minutes and that into 60 seconds which is to divide it into!
!
!
The smallest primes, smallest factors down to which we can reduce a fraction are, 2 and 3. We
have!
, !
And there are 18 groups in the periodic table of elements because the properties of the
element are periodic over cycles of 18. We have!
, !
, !
!
Is the golden ratio.!
And degrees come from in circle giving!
!
Is unity.!
!
!
!
!
(24hours)(60minutes)(60second s) = 86400secon ds
1Rotation
86400s
=
1Day
86400s
2 ⋅ 3 = 6
3 ⋅ 6 = 18
86400
2
= 43200
86400
3
= 28800
86400
12
= 7200
86400
6
= 14400
7200
100
=
7200
10
2
= 72
2 cos(72
∘
) = Φ =
5 + 1
2
360
∘
86400
360
= 240
2 cos(240
∘
) = − 1
86400
6
= 14400
14400
8
= 1800
86400
8
= 10800
6
8
=
3
4
= 0.75
of 48 70
So we are dealing with the reconciliation of the six-fold with the eight-fold. This suggests to me
the periodic table of the elements, and I had found something in earlier work that I had done
this is group 14 in the periodic table of elements its first two elements carbon, the core element
of life and silicon (Si) the core element of Earth (or sand) which is silicon dioxide ( ). Let’s
look at this on the next page….!
SiO
2
of 49 70
We see that the atomic radius of silicon the core element of artificial intelligence (transistor
technology) fits together with the core element of biological life carbon if the silicon is taken as
inscribed in a regular dodecagon (12 sides) and the carbon is taken as inscribed in a regular
octagon (8 sides). We have:!
, , !
, , !
Apothem: "
For a regular dodecagon:!
!
The radius of a silicon atom is Si=0.118nm and that of carbon is
C=0.077nm:!
!
!
!
This has an accuracy %!
Twelve is an expression of six in that it allows you describe it as two three’s and three two’s."
D = 1 + 2x
x
2
+ x
2
= 1
2
2x
2
= 1
2x
2
= 1
x = 2/2
D = 1 + 2
a = (1 + 2)/2 = 1.2071
a =
s/2
tan(θ /2)
=
0.5
tan(15
∘
)
= 1.866
Si
C
=
0.118
0.077
= 1.532
a
S
i
a
C
=
1.866
1.2071
= 1.54585
1.532
1.54585
100 = 99
of 50 70
11.0 The Second Is The Metric
We have !
11.1. !
11.2.
, and . In both 11.1 and 11.2 we have
kinetic energy over kinetic energy times one second equal to a pivotal time period in Nature, the
first the Life Span of the Universe, and the second the Earth Day, period of rotation of the Earth.
In equation 11.1 is given by the white dwarf star:
And
Is the Chandrasekhar limit for the mass of a star to not collapse into a blackhole so it can
become a white dwarf star. The Life Span of the Universe is perhaps related to the Schwartzchild
radius which gives the event horizon of a blackhole star. It is
11.3.
If is the mass of the Universe
11.4.
Where is the radius of the Universe and is the critical density of the Universe that
determines whether the universe is open, flat, or closed. It is
11.5.
Where
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
(1secon d ) = Li feSpanUniverse
K E
Earth
K E
moon
(1secon d ) = Ear th Day
h
G
⋅
c
3
m
p
= Kinet icEnerg y
m
i
(
1
k
)
2
= Kinet icEnerg y
m
i
m
i
= Mm
p
M = 0.77
c
3
h
3
8π
3
G
3
m
4
p
R
s
=
2GM
c
2
M
M =
4
3
π R
3
ρ
c
R
ρ
c
ρ
c
=
3H
2
0
8π G
of 51 70
11.6.
Is the radius of the Hubble sphere is observable edge of the Universe. is the Hubble constant
is the expansion rate of the Universe which is thought to be approximately 71km/s/Mpc. And,
the critical density is approximately the average density of the Universe. We can write 11.1
11.7.
Using where .
We have
EarthDay=(24)(60)(60)=86400 seconds
LifeSpanUniverse=1E14 Years = 3.734E21 seconds
KE Moon=3.428E28J
KE Earth=2.7396E33J
Kinetic energies used from aphelions and perihelions calculated earlier in this paper. We have
=
Using our honed value for k.
=(5.354E-34)(1.5657E-10)=8.3827578E-44
Multiplying these last two:
(3.96631kg)(8.3827578E-44)=3.32486E30 kg
This mass should equal that of a white dwarf star (Its lower limit not to collapse into a
blackhole). There can be some play as we are suggesting the moon is an indicator somehow of all
that is going on in the Universe, and it only has to be at approximately the right orbit to do this.
We have
Which makes the equation pretty accurate in this form considering we are comparing something
like the Earth day to something as immense as the LifeSpanUniverse. We are going to want to
look at:
c
H
0
= R
HS
H
0
ρ
c
ρ
k
4
m
3
p
⋅
h
2
c
6
G
2
(
Ear th Day
Li feSpanUniverse
)
2
(
K E
moon
K E
earth
)
2
= M
WhiteDwa r f
m
i
= Mm
p
M = M
WhiteDwa r f
k
4
m
3
p
⋅
h
2
c
6
G
2
=
(
1
788.4626
)
4
1
(1.6726E − 27)
3
(6.626E − 34)
(6.674E − 11)
2
(299792459)
6
(2.58747E − 12)(2.1371E80)(9.926581E − 47)(7.25979E50) = 3.96631k ilogr a m s
(
Ear th Day
Li feSpanUniverse
)
2
(
K E
moon
K E
earth
)
=
(
86400
3.734E 21
)
2
(
3.428E 28J
2.7396E 33J
)
2
2.86416E 30kg
3.32486E 30kg
= 86 %
of 52 70
Where is the Chandrasekhar limit. We now write
Where is the Schwartzchild radius is the is the edge of the observable universe. If
we have
Then
Yields
Which is the critical density of the Universe. We write 11.8
11.8.
h
G
⋅
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
M
c
8π G
3H
2
0
⋅
(
c
H
0
)
ρ
c
= R
s
R
s
c
H
0
= R
HS
R
HS
= R
S
8π G
3H
2
0
⋅
(
R
s
)
ρ
c
= R
s
8π G
3H
2
0
⋅ ρ
c
= 1
ρ
c
=
3H
2
0
8π G
h
G
⋅
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
H
0
= 71,000m /s /Mpc
(3.26E6ly /Mpc)(9.461E15m /l y) = 3.1E 22m /Mpc
H
0
= (71000m /s)(3.1E 22m) = 2.3E − 18s
−1
ρ
c
=
3(2.3E − 18)
2
8(3.141)(6.6741E − 11)
= 9.461E − 27kg/m
3
of 53 70
Thus we have
Where is a white dwarf star. We have
We suggest for some mass , we have
It is given by
11.9.
For all practical purposes this is the mass of the Moon, which is exactly 7.34767E22kg.
Conclusion next page…
h
G
⋅
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
M
c
K E
Earth
K E
moon
(1secon d ) = Ear th Day
M
h
G
⋅
c
3
m
p
M
(
1
k
)
2
(1secon d ) = 1Ear thYear
M =
h
G
⋅
c
3
m
p
(1Ear thYear)
(
1
k
)
2
(1secon d )
h
G
⋅
c
3
m
p
=
6.62607E − 27
6.67408E − 11
⋅
(299792459)
3
1.67262E − 27
= 1.599E 36J ≈ 1.6E 36J
1
k
2
= (788.4626)
2
= 621,673.27
1Ear thYear = (365.25)(24)(60)(60) = 31557600s
M =
1.599E 36
621673
⋅
1
31557600
= 8.15E 22kg
8.15E 22kg
7/34767E 22kg
= 1.10919516 ≈ 1.12m oon s
of 54 70
Thus we have
Where is a white dwarf star and gives LifeSpanUniverse.
Where Earth-Moon-Sun orbital parameters give the Earth rotation period.
Where is the mass of the moon gives the Earth orbital period. We have already formulated
From the radius of a proton and the mass of proton , our theory panned out: The second is
a unit of time that expresses the cycles of the Universe, it is the metric (measures thing
conveniently). I think next we want to look at
h
G
⋅
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
M
c
K E
Earth
K E
moon
(1secon d ) = Ear th Day
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
M
m
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
r
p
m
p
8π G
3H
2
0
⋅
(
c
H
0
)
ρ
c
= R
s
c
H
0
= R
HS
ρ
c
=
3H
2
0
8π G
of 55 70
12.0 Kilogram-Seconds
One of the things we have done is said since the Chandrasekhar limit can be taken as 1.44
solar masses, and the mass of the Sun is 1.98847E30kg, and the mass of a proton is
1.67262E-27kg, that our intermediary mass is!
!
We also computed!
Where came from the approximation of (3/4) for 0.77 in
Which we ultimately took as our value for the honed :
We have
We have honing our
We developed a concept of proton-seconds. Now we develop a concept of kilogram-seconds. To
do this we look at our equations for the cycles of the Universe:
12.1.
12.2.
m
i
= (1.44)(1.98847E 30kg)(1.67262E − 27) = 69.205kg
m
i
=
3
2
c
3
h
3
8π
3
G
3
m
2
p
1/2
= 67.9943kg
3/2
m
i
= Mm
p
= 0.77
c
3
ℏ
3
G
3
N
m
2
p
1/2
m
i
3
2
= 0.8660
0.77 = 0.8775
0.8775
0.8660
= 1.01328
m
i
m
i
= (67.9943)(1.01328) = 68.897kg ≈ 69kg ≈ 70kg
h
G
⋅
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
K E
Earth
K E
moon
(1secon d ) = Ear th Day
of 56 70
12.3.
12.4.
We consider
12.5.
12.6.
And write
12.7.
Which is
12.8.
Where we remember
12.9.
Can be taken as
12.10.
If cancels with the numerator.
12.11.
12.12.
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
h
G
⋅
c
3
m
p
=
6.62607E − 27
6.67408E − 11
⋅
(299792459)
3
1.67262E − 27
= 1.599E 36J ≈ 1.6E 36J
1
k
2
= (788.4626)
2
= 621,673.27
h
G
⋅
c
3
m
p
(
1
k
)
2
(1secon d ) = 2.5721E 30k ilogr a m ⋅ secon ds
h
G
⋅
c
3
m
p
⋅ t
1
⋅ k
2
= 2.5721E 30kg ⋅ s
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= proton ⋅ secon ds
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
m
p
h
G
⋅
c
3
m
p
⋅
1
6α
2
⋅
r
p
m
p
⋅ k
2
h 4π
Gc
= 2.5721E 30k ilogra m ⋅ second s
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= proton ⋅ secon ds
of 57 70
We can now take 12.8 and divide by the mass of the moon:
(365.25)(24)(60)(60)=3157600 seconds is:
Meaning we have the Earth year to 90% accuracy. We can also divide by the mass of the Earth:
=
And we note as we said in the beginning of this paper the moon orbits the earth inclined 5
degrees to the Earth orbital plane (That is, to the ecliptic). In days the Earth goes about 5
degrees around the Sun because the earth goes through about 1 degree a day because there are
360 degrees in a circle and 365.25 days in a year. We can also divide kilogram-seconds by the
mass of the Sun to get:
And find it is very close to one second. We in fact have that if we use instead of aphelions and
perihelions to calculate the kinetic energies of the moon and earth, we can use their average
orbital velocities we have (Appendix 1):
Equation 12.13.
As opposed to
Equation 12.14.
Which allows us to write very accurately
h
G
⋅
c
3
m
p
⋅ t
1
⋅ k
2
= 2.5721E 30kg ⋅ s
2.5721E 30kg
M
moon
=
2.5721E 30kg ⋅ s
7.34767E 22kg
= 35005654.85s
31557600s
35005654.85
= 90 %
2.5721E 30kg
M
earth
=
2.5721E 30kg ⋅ s
5.972E 24kg
= 430693.2351s
430693.235s
31557600s
= 0.013647845years
73.27
c ycles
year
365.25d a ys
73.27
= 4.98498d a ys ≈ 5d a ys
2.5721E 30kg
M
⊙
=
2.5721E 30kg ⋅ s
1.989E 30kg
= 1.29316s
K E
moon
K E
earth
(Ear th Day) = 1.2secon ds
K E
moon
K E
earth
(Ear th Day) = 1.08secon ds
of 58 70
Equation 12.15.
As we can see while kilogram-seconds deals with Natural cycles and periods
12.1.
12.2.
12.3.
12.4.
Proton-seconds are structured around the core element of life, carbon, and the most abundant
elements in life carbon and hydrogen and in the most abundant element in the Universe,
hydrogen, as founded on six-fold symmetry characterized in the hydrocarbons, the skeletons of
biological life and organic chemistry:
Eq 2.5. !
Eq 2.6.
h
G
⋅
c
3
m
p
⋅
t
1
M
⊙
⋅ k
2
=
K E
moon
K E
earth
(Ear th Day)
h
G
⋅
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
K E
Earth
K E
moon
(1secon d ) = Ear th Day
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
1
α
2
m
p
h4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h4π r
2
p
Gc
= 1proton ⋅ 6secon d s = hydrogen(H )
of 59 70
13.0 Apophis
The reason that motivated me to write this paper was explained in the opening section, but not
in its more specific details. It was explained that the Moon seems to give us a hint that we are
here for a reason, like it perfectly eclipses the Sun, and that may have been a message to our
ancient ancestors, as well as to us today. Or in that the Moon holds the Earth’s inclination to
the ecliptic, that keeps the climate right to support life. These things were already known, but I
discovered something that made me pursue this as a paper, and that was in that both the
proton and the Moon define that duration of a second nearly perfectly, which I have found in
many ways make it a good unit for measuring Nature accurately. I call it a “Natural Metric”. In
light of how we got the second makes this an even deeper mystery; We got it from the way the
ancients divided up the rotation of the Earth and the Earth’s orbit to make a calendar that
works well. But I pursued this paper for an even deeper reason than mentioned: I was
wondering in light of negative forces in the world and positive forces in the world whether or
not we were slated for success as a species from the outset, and the Moon in light of what we
already knew, and what I found, gave me good reason to pursue the project. It panned out so
well that I began to feel there was something more than certainly going on here. Among other
things, like a theory for reality, I found what seems to be life cycles of the Earth and Universe. I
found!
Equation 12.15.
12.1.
12.2.
12.3.
12.4.
Eq 2.5. !
Eq 2.6.
h
G
⋅
c
3
m
p
⋅
t
1
M
⊙
⋅ k
2
=
K E
moon
K E
earth
(Ear th Day)
h
G
⋅
c
3
m
p
M
c
m
p
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
K E
Earth
K E
moon
(1secon d ) = Ear th Day
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
1
α
2
m
p
h4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h4π r
2
p
Gc
= 1proton ⋅ 6secon d s = hydrogen(H )
of 60 70
I decided to apply this to Apophis, an asteroid we thought might hit Earth in 2029 that could
have either wreaked havoc, or even brought about our extinction. Hence its name from the
Greek meaning “mayhem” and “chaos”. It is going to be in 2023 the nearest miss in history,
coming between us and the Moon and even within the orbits of our geosynchronous satellites. It
has an orbit of about year and comes near to Earth every seven years. We want to see if the
Moon predicts such an asteroid (of 7 years) with equation 12.3. This gives
The mass of the Moon is
But is predicted by equation 12.3 to be 8.15E22kg. Thus we divide this by our answer above for
M:
This may be the 7 years, where the year is one orbit of the Earth and as well approximately that
of Apophis. Much more needs to be done, but we seem to be doing well with these equations of
Natural cycles so far. Of course with the second as a Natural Metric, the year is Natural as well
so the easiest way to do this is in equation 12.3 since one lunar mass gives 1 Earth year, so will
the 7 years of Apophis near miss give seven lunar masses. But considering the system of
equations given above, the question becomes why 7 years gives 7 lunar masses. We have to solve
this system of equation above, further. We only know Apophis passes between the Earth and the
Moon in 2023. In this paper I offer it up with more editing and correction of typos, and the
above suggestion of looking further into the system of equations for Apophis. This makes us
suggest mass is inversely proportional to time:
M =
h
G
⋅
c
3
m
p
(1Ear thYear)
(
1
k
)
2
(1secon d )
h
G
⋅
c
3
m
p
=
6.62607E − 27
6.67408E − 11
⋅
(299792459)
3
1.67262E − 27
= 1.599E 36J ≈ 1.6E 36J
1
k
2
= (788.4626)
2
= 621,673.27
1Ear thYear = (365.25)(24)(60)(60) = 31557600s
7Ear thYear = 220903200s
(1.599E 36J )
(220903200s)(621673.27)
= 1.16435E 22kg
M
m
= 7.34767E 22kg
Apophis =
8.15E 22
1.16435E 22kg
= 6.9996 ≈ 7
T =
1
M
of 61 70
Where T is in Earth years and M is in Lunar masses. This is a good clue for solving the framing
of our theory further that the Moon is the key to the Universe. And we are going to to that now
in the next section.
of 62 70
14.0 Solving The System We now undertake since the second is Natural, to show the Earth
year is as well because the Earth goes around the Sun not just in year, but in a second in
another sense. That is we can make a variable and say:!
Equation 14,0 !
We start with!
Equation 14.1
We write
Equation 14.2
Equation 14.3
Where is the mass of the Earth, and is the mass of the Moon. We now substitute for
the mass the mass of Jupiter because it significantly carries most of the mass of the solar
system excluding the Sun. It is 1.899E27 kg. We have
Equation 14.5.
This is about 1 Earth day because 1 Earth day=86400 seconds. We want to suggest that in our
scenario that the Natural cycles in time are inversely proportional to mass:
Equation 14.6
Because if , then . If the equation is:
Equation 14.7
Where M and T are in lunar masses and Earth years.
Let . We have the mass of Jupiter in Lunar masses is
t
1
t
1
= (1secon d,1second /1year, . . . )
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
t
1
⋅
h
G
⋅
c
3
m
p
⋅ k
2
= 2.5721E 30kg ⋅ s
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear ⋅
M
e
M
m
= 2.1E 32kg ⋅ s
M
e
M
m
M
e
M
j
t
1
⋅
1
M
j
⋅
h
G
⋅
c
3
m
p
⋅ k
2
⋅
M
e
M
m
=
2.1E 32kg ⋅ s
1.899E 27kg
= 110584.5secon d s
1
M
∝ T
M = M
m
1
M
m
=
1
1
= 1year = T
M
e
M
m
⋅
1
M
= T
M = M
J
M
j
=
1.899E 27kg
7.34767E 22kg
= 2.5845E4Lu n arMa sses
of 63 70
This is about one day since there are 365.25 days in a year. We already said
And we have
Is about one day because there are 86400 days in a year. Thus we equate the latter two:
But, the expression on the left gives its answer in Earth years and the expression on the right
gives the answer in seconds. Let convert second to Earth years. Let
We have
So since the Earth orbital period and
We have
Again we have, but we divide by to put it in lunar masses on the left and let
:
Equation 14.8.
M
e
M
m
⋅
1
M
j
=
81.27
2.5845E4
= 0.00314years
t
1
⋅
1
M
j
⋅
h
G
⋅
c
3
m
p
⋅ k
2
⋅
M
e
M
m
=
2.1E 32kg ⋅ s
1.899E 27kg
= 1.2799d a ys
M
e
M
m
⋅
1
M
j
=
81.27
2.5845E4
= 1.15d a ys
M
e
M
m
⋅
1
M
j
= t
1
⋅
1
M
j
⋅
h
G
⋅
c
3
m
p
⋅ k
2
⋅
M
e
M
m
t
1
t
1
= (365.25)(24)(60)(60) = (1secon d )/31557600secon d s /year
t
1
⋅
1
M
j
⋅
h
G
⋅
c
3
m
p
⋅ k
2
⋅
M
e
M
m
= 110584.5secon d s
110584.5s
31557600s
= 0.00350years
31557600 = T
e
M
e
M
m
⋅
1
M
j
= 0.00314years
0.00350years ≈ 0.00314years
M
j
M
e
t
1
= 1secon d /T
e
M
e
M
m
⋅
1
M
j
/M
e
=
1secon d
T
e
⋅
1
M
j
⋅
h
G
⋅
c
3
m
p
⋅ k
2
⋅
M
e
M
m
of 64 70
This is interesting because , is not a second but the Earth orbital period in
seconds. We have
Equation 14.9
This says
Equation 14.10
Let’s verify it, we wrote
We have
The measured mass of the moon is 7.34767E22kg, We are close by
accuracy
Thus we see
Equation 10 is our original equation, Equation 2.
We can write it
But now
t
1
= 1secon d /T
e
1 =
h
G
⋅
c
3
m
p
⋅ k
2
⋅
1secon d
M
m
T
e
M
m
=
h
G
⋅
c
3
m
p
⋅ k
2
⋅
1secon d
T
e
t
1
⋅
h
G
⋅
c
3
m
p
⋅ k
2
= 2.5721E 30kg ⋅ s
M
m
=
2.5721E 30kg
31557600s
= 8.15E 22kg
7.34767E 22kg
8.15E 22kg
= 90 %
t
1
= 1secon d,1secon d /year, . . .
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
⋅ t
1
= 1Ear thYear
t
1
= 1secon d,1secon d /1year, . . .
of 65 70
Because in lunar masses and Earth years it is to say
Since
Equation 14.11
Because
Equation 14.12
In the process we found out that
Where the mass of Jupiter gives one Earth day is one rotation of the Earth. We already had that
the mass of the Moon gives one orbit of the Earth:
We see the Solar system is like an atom, which is miraculously quantized in terms of the second,
which describes the atom and the Moon:
1
M
= T
1 =
h
G
⋅
c
3
m
p
⋅ k
2
⋅
1secon d
M
m
T
e
h
G
⋅
c
3
m
p
⋅ k
2
⋅
1second
M
m
T
e
M
= T
h
G
⋅
c
3
m
p
⋅ k
2
⋅
secon d
M
m
T
e
=
2.5721E 30kg
(7.34767E 22)(31557600s)
= 1.109 ≈ 1
t
1
⋅
1
M
j
⋅
h
G
⋅
c
3
m
p
⋅ k
2
⋅
M
e
M
m
= 1Ear th Day
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
K E
Earth
K E
moon
(1secon d ) = Ear th Day
of 66 70
15.0 Mars and the Moon Interestingly we can speak of the day of Mars and the day of the
Earth interchangeably because they are about the same, about 24 hour long for their rotations.
We have to keep in mind that Mars is smaller than Earth regardless that it completes a rotation
in the same time. The radius of Mars is 3,389,500 meters and that of the Earth is 6,378,100
meters. That is at its surface Mars rotates through 246.5 m/s and at the surface of the Earth
463.8 m/s.!
But interestingly, this ratio in size of Earth to the size of Mars is!
Equation 15.1 !
And the ratio of the Mars year to the Earth year is the same!
Equation 15.2. !
Interesting because the size of a planet has nothing to do with its orbital period. Orbital period
is a function of the distance of the planet from the star, not of its size. You could have an
asteroid at Mars orbit much smaller than Mars. The mass of Mars, of course is much lower too.
The mass of Mars is 6.39E23 kg and that of Earth is 5.97219E24 kg giving:!
Equation 15.3. !
The Earth is then about 9 times more massive than Mars. Equations 15.1 and 15.2 are to say!
Equation 15.4. !
Where is the Earth radius and is the Mars radius, and is the Earth year and is the
Mars year. This is interesting because it is similar to the parameters of the moon with respect
to the Earth and Sun that we suggest is a mystery that is a message from unknown forces. We
had in equation 3.3 for this!
Equation 15.5. !
Where is the earth orbital radius, is the Moon’s orbital radius, is the Sun’s
radius, and is the radius of the Moon. Indeed the mystery of the Moon could speak to us
of the mystery of our existence even to our ancient ancestors flaking spearpoints from stone
around the fire a million years ago because they could witness the Moon perfectly eclipsing the
Sun, as much as it could be a message to us today. But in 15.4 where the Earth size to the
Mars size equals the Mars year (journey across the sky) to the Earth Year (journey of the Sun
across the sky) would require us to know the size of the Earth and the size of Mars, so if it is a
message like the lunar message, it would only be readable later in the human story."
6,378,100
3,389,500
= 1.88
687
365.25
= 1.88
5.97219E 24
6.39E 23kg
= 9.34615
R
earth
R
mars
=
T
mars
T
earth
R
e
R
m
T
m
T
m
r
earth
r
moon
=
R
⊙
R
moon
r
earth
r
moon
R
⊙
R
moon
of 67 70
It would seem civilization on Earth was wise in ancient times to make a 360 degree circle,
because there are 365.25 days in a year (one orbit of Earth around the Sun) and this is close to
360 degrees, meaning as such the Earth moves through about one degree a day. 360 degrees
divides nice (evenly) by the angles in special triangles, the 60-60-60 and 30-60-90 and
45-45-90. These are the equilateral triangle, the equilateral triangle with its altitude drawn in,
and the square with its diagonal drawn in. The equilateral triangle, and the regular hexagon
made of six equilateral triangles, and the square, are the regular tessellating polygons, that is
they can tile a surface without leaving gaps. (See illustration ).!
But a civilization that arose on Mars would do it differently because the orbital period of Mars
gives it a 687 day year. Interestingly we can speak of the day of Mars and the day of the Earth
interchangeably because they are about the same, about 24 hours long for their rotations. But
since the Mars year is 687 days, one might guess that a civilization on that planet would divide
the circle into 690 degrees, so it would be close to the 687 day year, but still be mathematically
functional like 360 degrees is with the Earth.!
We see the 360 day year factor into the primes!
!
The 365.25 day year has 24 for each day, each hour of 60 minutes, and each minute of 60
degrees. We got this from the ancient Sumerians, the Babylonians, and the ancient Greeks. It
was their Sexagesimal, or base 60 counting. It is believed they had it because 60 is evenly
divisible by so much that is key:!
!
Which corresponds to !
!
Respectively. We ask if the Mars year is 687 days where Earth days and Mars days are the
same duration, and it is divided up into 690 degrees, what would a civilization make for a
second? Its day is 24 hours like the that of the Earth, would it be divided up into 24 units, and
then, from there into 60 units, and from there 60 more, using sexagesimal like inherited from
the Ancients? It might be the case.!
We might guess the Mars hour would come from dividing the Mars day into 30 units, because
30 divides evenly into 690, and 24 does not. Each of those hours might be divided into 60
minutes because 30 divides evenly into 60. And finally seconds would be 60 per minute
because that divides even as well into 60 as 1. So the total number of the Martian seconds in a
year would be!
!
The seconds in the Earth year are!
!
Thus!
!
2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 5 = 360
(2,3,4,5,6,10,12,15)
(30,20,15,12,10,6,5,2)
(687)(30)(60)(60) = 74196000secon d s /MarsYear
(365.25)(24)(60)(60) = 31557600secon d s /EarthYear
74196000s/687da ys = 108000s /day
of 68 70
!
There are!
!
Thus there would be 1.25 Mars seconds in 1 Earth second, is to say there are more of them
because they are 0.8 Earth seconds long. Thus like this in one day, Mars goes around the Sun!
!
!
31557600s/365.25da y = 86400s /da y
108000
86400
= 1.25
690/687 = 1.0043668
∘
≈ 1
∘
of 69 70
Appendix 1
We suggested the second was a natural unit, and that, if it was, should be in the orbital
mechanics of the earth because the second comes from the calendar, which is based on the
orbital period of the year (1 year, 365.25 days) and the orbital period of the moon, and the
rotation of the Earth. We found that it was as the following:
Let’s show that…
To find the translational kinetic energy of the moon:
Distance from earth: 3.85E8m
Orbital period:
T=27.32 days=2.36E6seconds
v=1.025E3m/s~1000m/s
Mass: 7.34767E22kg
Use
E=3.67E28 Joules
To find the translational kinetic energy of the earth:
Distance from Sun: 1AU=1.496E11m
Orbital period: 1 year=3.1558E7 seconds
v=2.9785E4m/s, Earth mass: 5.9722E24kg
E=2.649E33 Joules
Earth day=(24)(60)(60)=86,400 seconds."
K E
moon
K E
earth
(Ear th Day) ≈ 1secon d
K . E . Moon = 3.67E 28J
K . E . Ear th = 2.649E 33j
2π (3.85E8m) = 2.419E 9m
E =
1
2
mv
2
2π (1.496E11m) = 9.399E11m
of 70 70
The Author!